Let $R$ be the region enclosed by the line $y=1$, the line $x=4$, and the curve $y=\dfrac 1x-1$. $y$ $x$ ${y=\dfrac 1x-1}$ ${y=1}$ $x=5}$ $\left(\dfrac 12,1\right)$ $(4,1)$ $\left(4,-\dfrac34\right)$ $ R$ A solid is generated by rotating $R$ about the line $x=5$. Which one of the definite integrals gives the volume of the solid? Choose 1 answer: Choose 1 answer: (Choice A) A $\pi\int_{\frac 12}^4 \left( 15+\dfrac{2}{y}-\dfrac{1}{y^2}\right)dy$ (Choice B) B $\pi\int_{\frac 12}^4 \left( 24-\dfrac{10}{y+1}+\dfrac{1}{(y+1)^2} \right)dy$ (Choice C) C $\pi\int_{-\frac 34}^1 \left( 15+\dfrac{2}{y}-\dfrac{1}{y^2} \right)dy$ (Choice D) D $\pi\int_{-\frac 34}^1 \left( 24-\dfrac{10}{y+1}+\dfrac{1}{(y+1)^2} \right)dy$
Explanation: Let's imagine the solid is made out of many thin slices. Each slice is a cylinder with a hole in the middle, much like a washer. $y$ $x$ ${y=\dfrac 1x-1}$ ${y=1}$ $x=5}$ $\left(\dfrac 12,1\right)$ $(4,1)$ $\left(4,-\dfrac34\right)$ Let the thickness of each slice be $dy$, let the radius of the washer, as a function of $y$, be $r_1(y)$, and let the radius of the hole, as a function of $y$, be $r_2(y)$. Then, the volume of each slice is $\pi[(r_1(y))^2-(r_2(y))^2]\,dy$, and we can sum the volumes of infinitely many such slices with an infinitely small thickness using a definite integral: $\int_a^b \pi [(r_1(y))^2-(r_2(y))^2]\,dy$ This is called the washer method. What we now need is to figure out the expressions of $r_1(y)$ and $r_2(y)$, and the interval of integration. $r_1(y)$ is equal to the distance between the curve $y=\dfrac 1x-1$ and the line $x=5$. To find it, we need to solve the equation for $x$ : $x=\dfrac{1}{y+1}$ So, ${r_1(y)=5-\dfrac{1}{y+1}}$. $r_2(y)$ is equal to the distance between the line $x=4$ and the line $x=5$. So, ${r_2(y)=1}$. Now we can find an expression for the area of the washer's base: $\begin{aligned} &\phantom{=} \pi [({r_1(y)})^2-({r_2(y)})^2] \\\\ &= \pi [\left({5-\dfrac{1}{y+1}}\right)^2-({1})^2] \\\\ &= \pi\left( 25-\dfrac{10}{y+1}+\dfrac{1}{(y+1)^2}-1 \right) \\\\ &= \pi\left( 24-\dfrac{10}{y+1}+\dfrac{1}{(y+1)^2} \right) \end{aligned}$ The bottom endpoint of $R$ is at $y=-\dfrac 34$ and the top endpoint is at $y=1$. So the interval of integration is $\left[-\dfrac 34,1\right]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_{-\frac 34}^1 \pi\left( 24-\dfrac{10}{y+1}+\dfrac{1}{(y+1)^2} \right) dy \\\\ &=\pi\int_{-\frac 34}^1 \left( 24-\dfrac{10}{y+1}+\dfrac{1}{(y+1)^2} \right)dy \end{aligned}$